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Haus, Hermann A., and James R. Melcher. Solutions Manual for Electromagnetic Fields and Energy. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-NonCommercial-Share Alike. Also available from Prentice-Hall: Englewood Cliffs, NJ, 1990. ISBN: 9780132489805. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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SOLUTIONS TO CHAPTER 13

13.1 INTRODUCTION TO TEM WAVES

13.1.1 (a) From (13.1.3):

aEz [ () (O)J aHaay = f3Re A cos f3y exp 3wt = Wat: (1)

= f3IAI cosf3ycos(wt +~)

where ~ is the phase angle of A. Integrating the above yields

H. = LIAlcosf3ysin(wt +~) = -Re jLAcosf3ye;wt (2) w~· w~

Introducing (2) and the expression for E z into (13.1.2) gives

- f32 IAI sin f3y sin(wt + ~) = -wflAI sin f3y sin(wt + ~) (3) w~

from which the dispersion relation follows f32 = W2~f.

(b) From (13.1.13)

This gives, using (2),

(4)

and thus

A= _j w~/(o = -j/(o. ~_1_ (5)f3 cos f3b V-; cos f3b

Using (2) we find

- -R K'" cos f3y ;wtH• - e 0 cos",Qb e (6)

and putting the value of A from (5) into the expression for Ez gives

= -R oK f;~ sin f3y ;wtEz e 3 0 f cos", Qb e (7)

1

Solutions to Chapter 13

13.1.2 (a) The standing wave H. = Re A sin {iyeiwt

satisfies the boundary conditions of sero H. at y = O. From (13.1.2)

aHa . t aE:I'-- ={iRe A cos {iye'W =E-- (1) ay at

Integrating to find Ell' gives

E:I' = -!!...Re jA cos {iyeiwt (2)WE

From (13.1.3) we find

aE:I' = {i2 Re jAsin{iyeiwt = J.& aHa = wJ.&Re jA sin {iyeiwt (3) ay WE at

and thus (i2 = W2 J.&E (4)

(b) Turning to the boundary conditions,

E:I'(-b, t) = Re ~deiwt /a (5)

and thus from (2)

-!!...Re jAcos{ibeiwt = Re ~deiwt /a (6)WE

and hence A- .WEVd _1__ . ~Vd_1_ (7)-3 (i a cos{ib -3yp. a cos{ib

We find

13.1.3 Using the identity (1)

one finds from (13.1.17)

l'liIE:I' = "'!¥ 1 •-Rej.n.o -----;(e'''I- e-'''")e'Wiii 't

Ecos{ib 23 (2)

= -Re !Ko ~[ei(wt-{JII) - e-i(wHfJlI)J/cos{ib2 y;

The exponentials in the brackets represent waves that retain constant amplitude when dy = ±idt exhibiting the (phase) velocities ±w/{i = ±1/..,fiii.

13-3 Solutions to Chapter 13

-L/ 13.1.4 (a) The EQS potential in a coax is a solution of Laplace's equation. The field

with rotational symmetry is

~= Alnr

(I)a

satisfying ~ = 0 on outer conductor of radius a. The field is z-independent with a constant potential difference. The potential difference is

Aln(b/a} =V (2)

The field is

E = -V~ = -i.. :rA1n(r/a} = -i.. ~ = i .. rln~/b} (3)

(b) The field has cylindrical symmetry with field-lines parallel to ill>. The potential "\Ii' is

(4)

The H field is

(5)

Ampere's integral law gives

!H.dS= f J·da=I (6)

Since H is z independent, I = constant and at z = -I

A --211"r = -211"A = I (7)

r

Therefore

(8)

(c) The preceding analysis suggests that

E=i.. V(z,t} (9a)In(a/b}r

and

(9b)

can be solutions of Maxwell's equations. To show this it is advantageous to separate the V operator into

(10)

Solutions to Chapter 13 13-4

where T"7 • a 1. a vT = I r + -1",

ar r a</>

is the transverse part of the operator. Then

(ll)

v X E = V T X E + i. X :zE (12)

Now V T differentiates only rand </>. The EQS field, which is z independent, has VT X E = O. Hence we conclude that the same holds for the "Ansatz" (9). But i. X ir = i", and i. xi", = -ir . We obtain from Faraday's law

1 !~V=_JL_l_a1 In{a/b) r aa 21Tr at

(13)

The common r-dependence can be eliminated, and we find

(14)

where L = JLln{b/a)

21T A similar reasoning applied to V X H and Ampere's law yields

(15)

• 1 a1. € av -Ir =Ir -

21Tr az In{b/a)r at -- (16)

or

with

a1 = _cay az at

c = 21T€ In{b/a)

(17)

(18)

V 13.1.5 (a) With the time dependence exp iwt, we get for the transmission line equations

of (14) and (17) of Prob. 13.1.4

dV dz

= -iwLJ (I)

where

dJ A

- = -iwCVdz

v = Re Veiwt

(2)


and 1 = Re ieiwt

Eliminating V from (1) and (2) one obtains

tPV . di 2- = -3wL- = -w LCVA (3)

dz2 dz

with the solutions

(4)

with fi =w";LC (5)

We pick the solution v = Asinfiz (6)

because the short forces V to be zero at z = O. From (1) we find

i dV ifi1

A =-- = -Acosfiz (7)wL dz wL

and since 1 = Re 10 eiwt at z = -I,

wLAcosfil = -i 1 (8)

fi 0

or

A=-iVL/C~ (9)cos PI

where we used (5). We find for the current and voltage as functions of z and t:

. t101(z, t) = Re -- cos pze'w (10)cosfil

V(z, t) = -Re iVL/C10 sin~z,eiwt (11) cOSfJ

(b) At low frequencies cosfiz!::::! 1 for all -I < z < 0 and sinfiz !::::! fiz = w";LCz. Using (9) of the preceding problem,

(12)

For the E-field we find from the preceding problem and (11) above

. R' L 1 iwt _. R . JJ 1 iwtz oeE -- -II' e3w In(a/b)r - -II' e 3w 211" z oe (13)

13-6

This gives the voltage at z = -I


(14)

The inductance is Ll because L, as defined here, is the inductance per unitlength. Thus we have shown that, in the limit oflow frequencies, the structurebehaves as a single-turn inductor.

(c) The H-field in the space between the conductors is the gradient of a potential'Ii ex tP that is a solution of Laplace's equation. Thus,

10 • tH=Re -i e'w

21fr .p

We obtain E from Faraday's law

V E paH R' 10 • iwtX =--- = -p e:Jw-1.peat 21rr

(15)

(16)

cI

f--- -----.-_.-t--

z « 0) z=o

FllPlre SII.1.5

With the line integral along the contour C shown Fig. S13.1.5, we may find fromthe integral form of Faraday's law

(17)

Integrals over the radial coordinate appear on both sides. Thus, comparing theintegrands we find

which is the same as (13).

E R jwp10 iwtr=- e~ze (18)


~ 13.1.6 (a) From the solutions (4) in Prob. 13.1.5 we pick the cospz dependence, because

the magnetic field, proportional to 1, is zero at z = 0 according to (7) of the same problem. Indeed, if V = A cos pz, then

j eW jP.1

A

= -- = --Asmpz (1)wL dz wL

Since Re[Acospzexpjwt]..=_1 = ReIVoexpjwt] (2)

we find

A=~ (3) cos Pi

and

1=-jyC/L V°/.llsinPz (4) cos~

Therefore,

V(z, t) = Re [c:opl cospzexPjwt] (5)

1(z, t) = -Re jyC/L V°/.l sin (3zeiwt (6) cos~l

In(r/a) • V(z, t) 1 E = V(z, t)VT In (a/b) = I"ln(a/b) r (7)

where VT is the transverse gradient operator,

• a . 1 aV T = 1.. - +1",-ar r a4J

and we use the result of Prob. 13.1.4. In a similar vein

(8)

(b) At low frequencies, cos (3z !:::! I, sin (3z !:::! (3z and V(z, t) !:::! Re Vo exp jwt. Then, assuming Vo to be real,

i.. 1 ( ) E = In(a/b) r Vocoswt

H = ;:r yC/L(3zVosinwt = i"'rln'(:/b)zVosinwt (10)

(c) At low frequencies, using EQS directly

i.. 1 E = In(a/b) r Vocoswt (11)

9


namely the gradient of a Laplacian potential ex In(r/a). The H-field follows from

aEVXH=f (12)at

with A

H = 14>-z (13) r

introduced into (12)

• a H I A II" 1V. .V x H = -II" az 4> = - 1"-;:- = -WE In(a/b) r oSlnwt

and therefore

A = In(:/b) Vosinwt (14)

which gives the same result as (10).

13.2 TWO-DIMENSIONAL MODES BETWEEN PARALLELPLATES

13.2.1 We can write

mr 1 ( .n", .n'll" )cos-:r; = - exp3-:r;+exP-3-:r;a 2 a a

and • n'll" 1 ( .n", .n'll" )

Sln-:C= --; exp3-:r;-exp-3-:r;a 23 a a

Introducing these expressions into (13.2.19)-(13.2.20) we find four terms of the form

'Q .n'll" '(Q n'll" )exp =f3fJnyexp =f3 -:r; = exp =f3 fJnY ± -:r; = exp -jk . r a a

where k n'll". Q •

= ±-Ix ±fJnl~ a

andr = Ix:r;+I~y

This proves the assertion that the solution consists of four waves of the stated nature. These waves are phased so as to yield :r;-dependences of the form cos n: :r; and sin nat!' :c to satisfy the boundary conditions.


13.2.2 We can start with the solutions (13.2.19) and (13.2.20) shifting z so that , a

:1;=:1;-2

Considering TM modes first we note that

n", .(n",z' n",)Hz ex cos-z = cos --+

a a 2 n"':I;' n", . n"':I;' . n",

=cos--cos- -sm--sma 2 a 2

= { (-1)~~ cos (m::) n even (-1)-~- sin (,,~S) n odd

We see that the modes with even n are even with respect to the symmetry plane of the guide, the modes with n-odd are odd.

Next studying the TE-modes,

. n", . (n",z') n", (n",z') . n",Ezexsm-z=sm -- cos-+cos -- sma a 2 a 2

= {(-1):'~1 sin~, n even (-1)-~- cos "~s n odd

We find that Ez is even for n odd, odd for n even.

(a) When z, = ±a/2 and the modes are odd, Hz = (_1)(,,-1)/2 sin "2ft , Ez = (-1)"/2 sin "2ft ; in the first case n is odd and Hz is an extremum at z' = ±a/2, and in the second case n is even and Ez is zero at both boundaries.

(b) When z, = ±a/2 and the modes are even then Hz = (-I)"/2cos(;ft) and Ez = (_1)("-1)/2 cos ;'11' we see that both boundary conditions are in both cases, because n is odd in the first case and Hz is an extrenum, n is even in the second case, and Ez is zero.

13.3 TE AND TM STANDING WAVES BETWEEN PARALLELPLATES

(1)

13.3.1


where we have integrated by parts. Because dh.n/dz = 0 at z = 0 and z = a, the integral of the integrand containing the total derivative vanishes.

Next take the complex conjugate of (13.3.1) applied to h.m multiply by 'h.n and integrate. The result is

13-10

(2)

Subtraction of (1) and (2) gives

Thus (G A. '" 10 h.mh.ndz = 0

when p~ -=I p~ and orthogonality is proven. The steps involving 2.n are identical. The only difference is that

la

d ~("* de.n )zd e.m do z z

vanishes because 2:m vanishes at z = 0 and z = a.

13.3.2 (a) The charge in the bottom plate is

q = l UI 1(a+A)/2 EElIdzdz (1)

o (a-A)/2

Using (13.3.15)

""' 4mrE f) 1 ;wt] 2wa ( )!!=.!.. n",~ q= Re [ LJ -------e -- -1 sm-- (2)3 ..=1 a fJn a sinfJnb n", 2a 044

lwhere we have used the fact that

UI j(a+A)/2 mr wa [ mr a + f1 mr a - b. ]sin (-z)dzdz = -- cos (---) - cos (---)

o (a-A)/2 a n", a 2 a 2

wa . n", . n",f1 = 2-sm-sm-

n", 2 2a _ 2wa ( 1).!!;.l . mrb. ---- ~ sm-

nll" 2a (3)

Va = -Re iwqejwtR

( 1) !!=.!. • n!tA. ] (4)= -Re iw8EWRv L - 3 s~ 2a ejwt

[ fJn a sm fJn b n


When P",b = 1r we have a resonance. Now

P", = JW2~E _ (n;)2

The resonance frequency of the n-th mode occurs at

p",!a = J4w2~Ea2 - (n1r)2 = 1ra

orw..jiii.a = J(n2 + 1)~

2

13-11

(5)

(6)

(7)

(b)

(b) For n = 1 this is at 1r. The next mode resonates at V5f41r. Thus, in thisrange, two resonances occur for which the response goes to infinity. Of course,in this limit, losses have to be taken into account which will maintain theresponse finite. The low frequency limit is when·

1rw..jiii. <: n

a

ThenR • R b n1r. h n1r bf'''' Slnf'''' -+ -- sm -

a aand

_ R [. R"~ (-1)!!jl sin~ jwt]Va - e 3w81rEW V L..J M sinh Mb e

'" G G

(c) From (13.3.13), when only one mode predominates,

H R [4iWEfJ cos p",y n1r] jwt

.!:::! e -R-- . R bCos- ef'",a sm f'", a

where n = 1 at wy'iifa = 1r and n = 2 at wy'iifa = V5f41r. To get afinite answer, we need v/ sin p",b to remain finite as the resonance frequencyis approached.

13.3.3 (a) H. at x = 0 and x = a gives the surface currents in the bottom and topelectrodes. Because the voltage sources push currents into the structure inopposite directions, the surface currents, and H., have to vanish at the symmetry plane.

The x-component of the E field can be found directly from (13.3.14), replacing the sinp",y/sinp",b by cosp",y/cosP",b to take into account the changedsymmetry of the field

E R [ ~ 4v cosp",y n1r] ,·wiII: = e L..J -- cos -x e

,,_1 a cos P",b aodd

Because 8H./8y = iWEEs we find

H - R [~ 4jWEfJ sin P",y n1r J jwt• - e L..J R R b cos "'J e

,,=1 f'",a cos f'''' a"odd


13.3.4 (a) The flux linkage>. is (1)

and the voltage is dH.

Va = IJA--;It (2)

(b) From (13.3.13) we find that liI.1 is a maximum for :& = 0 and :& = a.

(c) From the detailed expression (13.3.13), using (2)

_ -R [~4W2IJEUA_1_ n1r X ] iwt Va - e L.J Q • Q bcos e

"=1 IJn a sIn IJn a "odd

(d) The loop should lie in the 11 - z plane. Then it links Hz that is tangential to the bottom plate.

13.3.5 The Ez field is derivable from a potential that is a square wave as shown in Fig. S13.3.5. We have

(1)

v

-,,---:Or----'--x=o J \ x=a

/I-II tl4-x=T X - 2

FIKure SIS.S.1

and using orthogonality, multiplication of both sides by sin n; :& and integration gies

U4 a • n1r av [ (n1r a + d n1r a - d)]12-An = V sIn -zdz =-- cos ---) - cos (-- 2 ~ a n1r a 2 a 2

2

av . (n1r) . (n1rd)= 2-SIn - SIn n1r 2 2a

We find4v. (n1r) . (n1rd)An = -SIn - SIn n1l' 2 2a

We may adapt (13.3.13)-(13.3.15) for this case by replacing 4v/n1r by

. (n1r) . n1rd4v/ n1r SIn "2 sm 2a


~ 4jwdj • (n1r) . (n1r d) C08 fJnY n1r] iwtH• = Re [ L..J -- sm - sm - cos -z e .._1 fJn a 2 2a sin fJnb a odd

~ 4v. (n1r) . (n1rd)sinfJnY n1r] iwtE :I: = Re L..J -- sm - sm -- cos -z e[ .._1 a 2 2a sin fJnb a odd

~ 4n1r . (n1r) . (n1rd) cos fJnY . n1r ] iwtE = Re L..J --sm - sm -- sm-z e y [ ..=1 a 2 2a sinfJnb a

odd

13.3.6 In (13.3.5) we recognized that E. at Y = b must be the derivative of a potential that is a square wave. This, of course, is equivalent to the statement that E. possesses two impulse functions. In a similar manner, H y can be considered the derivative of a flux function f: pHydz. Note the analogy between (13.3.19) and (13.3.14). We may, therefore, adapt the expansion of P13.3.5 to this problem, because the flux function of Example 13.3.2 is the same as the potential of example 13.3.1. From (13.3.17)-(13.3.19):

~ 4jAw. (n1r) . (n1rd) sinfJmY . m1r ] iwtE • = Re [ L..J ---sm - sm -- sm-z e m=l m1r 2 2a sin fJmb a odd

~ 4fJm A . (n1r) . (n1rd)cos fJmY . m1r ] ,·wtH• = Re L..J -- sm - sm - sm -z e[ m=l pm1r 2 2a sin fJmb a odd

~ 4A. (n1r) . (n1r d) sinfJmY m7l'] ,·wt1l = Re [ L..J --sm - sm - cos-z e y m=l pa 2 2a sinfJmb a

odd

13.4 RECTANGULAR WAVEGUIDE MODES

13.4..1 The loop in the Y - z plane produces H-field lines along the z-direction. IT placed in the center of the waveguide, at z = a/2, these field lines have the same symmetry as those of the TEIO mode and thus excite this mode. The detection loop links these fields lines as well Of course, the position of tl.~ exciting loop must be displaced along Y by one quarter wavelength compared to the capacitive probe for maximum excitation.


13.4.2 The cutoff frequencies are given by

The dominant mode has n = 0 and thus it has the (lowest) cutoff frequency

:c Im=l,n=o = (;)

The higher order modes have cutoff frequencies

The cuttoff frequencies are in the ratio to that of the dominant mode:

TEol 1.33

TEll and TMll 1.66

T~o 2.0

T~l and TM~u 2.4

13.4.3 (a) TM-modes have all three E-field components. They approach the quasistatic fields of Ex. 5.10.1 which imposes the same boundary conditions as this exam

ple. Hence the modes are TM. From (9) we find that ez oc -jle,/;: = ;;~~ .. ·Ie ~ ,,3al! S· ~ d ~ . h to;' tand e. oc -1 11 ". = "11 .' mce Gz an e. must van18 a y = ,e" mus

behave as a cosine function of y, so that 2z and 2. are sine functions of y. Therefore,

E = Re '""'(A+ e-:ilJm...1I + A- eilJm...") sin ~:z:sin ~ze·;wt" L.J L.J mn mn a w m n (1)

= Re EE2A~ncosPmnysinm1r :z:sin~:z:sin~zeiwt a a w

m n

where (2)

From (13.4.9):

Ez = Re "'''' -jPmn(7) (A+ e-ilJm..." - A- eilJm...")L.J L.J W2uE _ .02 mn mn m n ,.. f'mn

cos ~:z:sin ~zeiwt (3) a w

_ R '"'" fJmn 7- 2A+ . R m1r. n1r iwt- e L.JL.J 2 R2 mnsmf'mnycos-:z:sm-ze m n W I"E - f'mn a w


and similarly,

R "" "" 13m",~ A+ . II • m7f n1l" jwtE.= eL..JL..J 2 112 2 m",smpm",ysm-zcos-zem '" W /Sf - Pm", a w

13-15

(4)

(b) At y = b, Es as a function of z must possess two equal and opposite unitimpulse functions of content v(t)/a to give the proper voltage drop at theedges. The integral of Es , - f; Esdz must be a square wave function ofamplitude v. The same holds With regard to the integral of E. with respectto z. In summary, Es and E. at y = b must be derivable from a potentialthat is a two-dimensional square wave with the Fourier expansion (5.10.15)(coIQ.pare 5.10.11):

00 00

4>(z,y) = Re L L__ 1 ,,==1

modd ,,"odd

x=o

16v . m7f • n1l" ,·wt--sm-zsm-zemn1l"2 a w

impulsefunction

(5)

impulsefunction

Thus, at y = b

X

F1cure 913.4.3

x=a

Comparison with (3) gives

m1l"/( 2 2) . 16v m1l"2Am",13m",- W /Sf - 13m", sm13m",b = --2-a mn1l" a

(7)

for m and n odd. This gives the quoted result. An analogous relation may beobtained for E. which yields the same result.

(c) The amplitudes go to infinity when sin 13m",b = 0 or

or

13.4.4

13-16

(d)


wVjii.a = 1rJm2 + (:n)2 + (ip)2

We have already used the fact that the distribution of Es and E. in the Y= bplane is the same as in the quasistatic case. The only difference lies in they-dependence which, for low frequencies gives the propagation constant

and is pure imaginary. The EQS solution according to (5.10.11) and (5.10.15)IS

.". = Re ~ ~ 16& sinh kmny . m1r • n1r iwt"Ii! L.J L.J --2 . hk b Sin -ZSIn -ze

m=l ..=1 mn1r sin mn a wodd odd

and gives for Es :

E• -_ -Re ~ ~ 16& (m1r)sinhkmnY m1r. n1r iwt- L.J L.J -- - cos-zsm-ze1 1

mn1r2 a sinh kmnb a wm. .",odd odd

This is the same expre88ion as the EQS result.

z=w

a-tl2

a+tl2

w+~

2

w-~

2

___...l--_..L-_.L...-__-'-_

x=o%=0

x =4/2 x=a

Flsure SIS.4.4.

The excitation produces a H,," It looks like TE-modes are going to satisfy all theboundary conditions. H" must be zero at Y= 0 and thus from (25) of text

00 00

H" = Re L L(O~ne-ilfm.." + O';neilfm..") cos C:1rz) cos C:: z)eiwt

m=On=O

= -Re L L 2iO~n sin PmnY cos (~z) cos (~z)eiwtm=On=O a w

(1)

At Y= b we must represent the two dimensional square-wave in the z-direction andin Fig. S13.4.4b in the z-direction as shown in Fig S13.4.4c.


a-d-2- z=o

w-d-2-

z=w

x=o a/2 x=a

Figure SIS.4.4b

We have, settingFigure SIS.4.4e

(2)

1w14

", '" m1l" (n1r (P1l") (q1l" 1L-L-Amncos(-X)cos -z)cos -x cos -z)dxdz= --Apq(aw)00 a w a w 4

w+oIil. A

1 2 i 2 p1l" q1l"=-Ho dz dxcos (-x) cos-zJ!!:A ..-I! a w

2 2

w±oIil. ~

1 2 12 p1l" q1l"+ Ho dz dxcos (-x) cos-z.!!!=A A a w

2 2

H o [. (P1l" a) . (P1l" a- d)]= - (7) (~) sm 72 - sm 7-2-

[. (q1l" W+ A) . (q1l" W- A)]sm --- -sm ---

w 2 w 2

Ho [. p1l" a+ d . p1l" a]+ (7)(~) sm 7-2- - sm 72

[. (q1l"w+A) . (q1l"w-A)sIn --- -sm ---

w 2 w 2

Ho [. (P1l" a)]= - (7)(~) sm 72

. (P1I" a - d) . (P1l" a + d) . (P1l" a ]-sIn --- -sm --- +sm --)a 2 a 2 a2

[. q1l"w+A . q1l"W-A]sm----sm---

w 2 w 2

Ho [. (P1l") . (P1l") (P1l")] q1l". q1l"A=-( )("1r) 2sm - -2sm - cos -A 2cos-sm--P!! .0.::.. 2 2 2a 2 2w

4 w

4Ho • (P1I") (q1l"). q1l"A [ P1I"]= - (p~)(~) sm "'2 COS "'2 sm 2w 1- cos 2a A

(3)We find that P must be odd and q must be even for a finite amplitude to result.

A = Ho (-l)P-l(-l)f- l [l- cos p1I" A]pq pq1l"2 2a (4)


The case q = 0 must be handled separately.

1 () H o [ • (P1f' a) . (P1f' a - d)]-A 0 aw = --- sm -- - sm --- w 2 p (p~) a 2 a 2

H 0 [ • P1f' ( a+ d) . (P1f' a)]+-- sm- -- -sm -- w (5)(7) a 2 a 2

2wHo [ P1f'] . P1f' = - (p~) 1- cos 2a sm"2

and thus Apo = Ho [1- cos P1f' .6](-1)P-l

P1f' 2a

From (13.4.7) and (13.4.8), one finds

'" '" 2;C;:;'nPmn (m:) . m1f' n1f' ;wtHz = Re LJLJ 2 p2 cospmnysm-xcos-ze (7)

m n W JJ.€ - mn a w

(8)

with C;:;'n expressed in terms of the Amn's by (2)

13.5 DIELECTRIC WAVEGUIDES: OPTICAL FIBERS

13.5.1 (a) To get an odd function of x for e", one uses the Ansatz

Ae-a",(z-d) d<x<oo e - Asink",z -d < x < d (1)'" - { sink",d _Aea",(z+d) -00 < x <-d

which has been adjusted so that e", is continuous at x = ±d. Since

(2)

and thus -azAe-a",(z-d)

k - _1_ k A cosk.. z (3)l/ - . z sink",d 'WJJ. { -azAea",(z+d)

kl/ and e", are continuous at x = d. The continuity of e", has already been established. From the continuity of hl/:

(4)


The cutoffs are at k",d = (2n - 1H (see Fig. S13.5.1a).

IIIII---1-. _

k",d (2n - 1)11"211" k d --+ ----

'" 2

Figure SI3.6.la

(c) When according to 13.5.3

k",d = JW2IJ f i - k~ = (2n - 1)~

and w goes to infinity, then kll must approach w.,fiifi asymptotically.

(d) See Fig. S13.5.1h (Fig. 6.4 from Waves and Fields in Optoelectronics,H. A. Haus, Prentice-Hall, 1984).

Asymptote ~ .. w \I~

1.00.5

2. -T·wv~•••---

Asymptote tJ = (oJ ",;;t;

0.0

50

10.0

1....

Figure SI3.5.lb


13.5.2 The antisymmetric mode comes in when

Q", 'II" -=0 and k",d= k", 2

and from (13.5.8)

or

or1 '11"/2 [E; c . ~ '11"/2

w = VJifid V(l- -:J = V7"dV -;; ";1- E/Ei

8 = 3 X 10 _1_ '11"/2 = 3.85 X 1010 10-2 yI2.5. /1 - --L

V 2.5

f = .!!!.... = 6.1 X 109 Hz 211"

13.5.3 (a) For TE modes

Ae-a.,(",-d) x>d e - A cos k.,'" or A sin k a'" -d < x < d'" - { cos k.,d sin k.,d

Aea.,(",+d) or - Aea.,(",+d) x< -d

where we have allowed for symmetric and antisymmetric modes. Continuity of ellS has been assured on both boundaries. The magnetic field follows from

h = _I_de", (2) y jwJ1. dx

and thus

_.!!.£ Ae-a.,(",-d) x>d ~ 1 k"'· kh =- -='A~ or -d < x < d (3)Y jw "'i cos k.,d

{ .!!.£Aea.,("'+d) or x< -d'"

Continuity of hy gives k",

-Q",

= -tank",d (4a)J1. J1.i


for even modes, and

for odd modes. Here

and thus, eliminating k",

as = J1c'tJ - W2

1-'E

kIlO = JW2lJiEi - k~

13·21

(4b)

(5)

(6)

and

(7)

(b) Cutoff occurs when as/kIlO = 0 and ksd is fixed. We find that when IJi isincreased above 1-', W must be lowered.

(c) The constitutive law (a) for symmetric modes has the graphic solution of Fig.13.5.2. The only change is the expression for as/kIlO but its ksd dependenceis qualitatively the samei as/ kIlO increases when IJi / I-' increases at constantw. This means that the intersection point moves to greater ksd values. k~

increases directly with increasing 1Ji/I-' according to (6) and decreases withincreasing kIlO' The intersection point of ksd does not change as fast, in particular, at high frequencies it does not move at all. Hence, the direct dependenceon IJi predominates, k" goes up and A decreases.

13.5.4 (a) The fields are now

:.r:>d-d <:.r: < d:.r:< -d

(1)

where we have allowed for both symmetric and antisymmetric solutions. Cointinuity of h. has been asured on both boundaries. Further,

SinceA 1 dhae =---

" iWE d:.r:

(2)

(3)

(4)


_!!a.Ae-a,.(.-d)1

2 = --E

" iw a1n _!..A 1l:.. or !..coa1l:••

E; aln1l:,.d (5){ E; coa1l:,.d

~Aea,.(.+d) or _ !!a.Aea,.(.+d) E

Continuity of 2" at z = ±d gives

(6a)

for even modes, and

(6b)

for odd modes. Further,

(7)

Thus (8)

and

(9)

(b) The cutoff frequencies are determined by k.d = m~ and a. =o. From (9)

or

mit opencourseware haus, hermann a., and james r. melcher. … · 2017. 1. 19. · solutions to...

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